Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R):
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:
Show Hint
- (A) is false, but (R) is true.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- (A) is true but (R) is false.
- Both (A) and (R) are true, but (R) is NOT the correct explanation of (A).
The Correct Option is C
Solution and Explanation
The problem deals with understanding the behavior of a simple pendulum under different gravitational conditions. The time period \( T \) of a simple pendulum is given by the formula:
\( T = 2\pi \sqrt{\frac{L}{g}} \)
where:
- \( L \) is the length of the pendulum.
- \( g \) is the acceleration due to gravity.
On Earth, the gravitational acceleration \( g \) is \( \frac{GM_e}{R_e^2} \), where \( G \) is the gravitational constant, \( M_e \) is the Earth's mass, and \( R_e \) is the Earth's radius.
For the other planet, with mass \( 4M_e \) and radius \( 2R_e \), the gravitational acceleration \( g' \) is:
\( g' = \frac{G \cdot 4M_e}{(2R_e)^2} = \frac{4GM_e}{4R_e^2} = \frac{GM_e}{R_e^2} = g \)
Thus, the time period \( T' \) on the planet is:
\( T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g}} = T \)
This confirms that the time period on both Earth and the planet is the same, supporting assertion (A).
Now, consider the reason (R). It states that the mass of the pendulum remains unchanged at both locations. While this is true, the mass of the pendulum does not affect the time period \( T \), as evidenced by the formula. Therefore, (R) does not explain (A).
Thus, the correct choice is: (A) is true but (R) is false.
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